The ends of the latus rectum of the conic $x^{2} + 10 x - 16 y + 25 = 0$ are

(3,-4)

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(-3,-4)

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(3,4),(-13,4)

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(5,-8),(-5,8)

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Solution

The correct option is C (3,4),(-13,4)
$(x + 5)^{2}$ =16y, compare it with $x^{2} = 4 a y$

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